solving simultaneous linear equations

7/27/2019 Solving Simultaneous Linear Equations

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Matrix MethodsThe solution to these sets of equations

can be solved using matrix methods. The

simultaneous solution of multipleequations finds its way in to manycommon engineering problems. In fact,modern structural engineering analysistechniques are ALL ABOUT solvingsystems of equations simultaneously.


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Matrix Methods Matrices - an organized way of presenting a set ofcoupled equations.

We seek a single unique solution that satisfies all theequations at the same time.

Consider the three coupled linear equations below:


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Matrix Methods Coupled because each equation has one or more termsin common with the others, , so that a changein one of these variables will affect more than oneequation.

Linear because each equation contains only first orderterms of . There are no terms like

Using the rules of matrix multiplication, we canrepresent the above equations in matrix form:


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Matrix Methods


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Matrix MethodsTry multiplying the matrices A and X

together, make sure you can get the original

equations above. There are several ways tosolve for unknown vector . Each methodinvolves some manipulations to thecoefficient matrix using algebraic rules,

creating a new and equivalent problem in amore easily solvable form. Thesemanipulations involve the addition of

multiples of one row to another.


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Matrix Methods


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Matrix Methods If we plot these 3 equations, the solution is the placewhere they intersect. That is, we are seeking the onepair of X1 and X2 values which lines along both or theoriginal lines (eq1, eq2).


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Gaussian Elimination


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Gaussian Elimination KEY:Whatever we do to the l.h.s. of an equation, wedo to the r.h.s. so we dont change the problem.

Lets rewrite the matrix in Eq. (4) to get one matrixwith both A and B in it:

This is called augmenting the matrix.


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Gaussian Elimination Step 1: - reduce A(2,1) to zero


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Gaussian Elimination If there were some numerical error in the computer storageof any coefficient, say the error from rounding off the -1/3currently in spot A(2,2), then when we multiply Row 2 by

some factor and add it to Row 3, we also multiply the errorby that factor.

If we can always multiply by some small number (less than1), we can reduce the propagation of that round-off error.

We can enforce this by making sure the lead coefficient(the pivot coefficient) in the pivot row has the largestabsolute value among itself and all the coefficients under it(the coefficients to be reduced to zero).


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Gaussian Elimination Since it does not matter what order I put the equationsin, we will rearrange rows when we find the currentpivot coefficient has a smaller absolute value thanthose beneath it. In the current example we have:


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Gauss-Jordan Elimination Gauss-Jordan Elimination - finding the inverse Gauss-Jordan Elimination I: For the first version of this

method, lets continue where we left off in the Gaussian

elimination example:

where Gaussian elimination is used to get the lowertriangle of zeros.


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Gauss-Jordan Elimination The idea of Gauss Jordan elimination is to continue thealgebra to get an upper triangle of zeros, until thethree equations are completely uncoupled.

Just as we worked from top to bottom to get zeros forA(2,1), A(3,1), and A(3,2) in that order, we can startworking from the bottom up to make A(2,3), A(1,3),and A(1,2) zero, in that order.

Just start with the bottom as the pivot row to zeroA(2,3), A(1,3), then switch the pivot row to the secondrow to make A(1,2) zero.


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Gauss-Jordan EliminationThe three steps to get rid of A(2,3), A(1,3),A(1,2) would be

Step 1) New Row 2 = (Row 3)(-11/6) + (Row2)

Step 2) New Row 1 = (Row 3)(1) + (Row 1)

Step 3) New Row 1 = (Row 2)(15/11) + (Row 1)


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Gauss-Jordan EliminationAnd the result is:


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Gauss-Jordan Elimination


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Gauss-Seidal Method

http://numericalmethods.eng.usf.edu

An iterativemethod.

Basic Procedure:

Algebraically solve each linear equation for xi

Assume an initial guess solution array

Solve for each xiand repeat

-Use absolute relative approximate error after each iteration

to check if error is within a pre-specified tolerance.


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Gauss-Seidal MethodAlgorithm

A set of nequations and nunknowns:

11313212111 ... bxaxaxaxa nn

2323222121 ... bxaxaxaxa n2n

nnnnnnn bxaxaxaxa ...332211

. .

. .

. .

If:the diagonal elements arenon-zero

Rewriteeach equation solvingfor the corresponding unknown

ex:

First equation, solve for x1

Second equation, solve for x2


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Gauss-Seidal MethodAlgorithmGeneral Form for any row i

.,,2,1,

1

nia

xac

xii

n

ijj

jiji

i

How or where can this equation be used?


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Gauss-Seidal MethodSolve for the unknowns

Assume an initial guess for [X]

n

-n

2

x

x

x

x

1

1

Use rewritten equations to solve for

each value of xi.

Important: Remember to use the

most recent value of xi. Which

means to apply values calculated to

the calculations remaining in the

currentiteration.


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Gauss-Seidal MethodGiven the system of equations

15x-3x12x321

283x5xx 321 7613x7x3x 321

1

0

1

3

2

1

x

x

x

With an initial guess of

The coefficient matrix is:

1373

351

5312

A

Will the solution converge using the

Gauss-Siedel method?


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Gauss-Seidal Method:

Example

1373

351

5312

A

Checking if the coefficient matrix is diagonally dominant

43155232122

aaa

10731313 323133 aaa

8531212 131211 aaa

The inequalities are all true and at least one row is strictlygreater than:

Therefore: The solution should converge using the Gauss-Siedel Method


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Gauss-Seidal Method :

Example

7628

1

1373351

5312

3

2

1

aa

a

Rewriting each equation

12

531 321

xxx

5

328 312

xxx

13

7376 213

xxx

With an initial guess of

10

1

3

2

1

xx

x

50000.0

12

150311

x

9000.45135.028

2

x

0923.3

13

9000.4750000.03763

x


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ExampleThe absolute relative approximate error%662.67100

50000.0

0000.150000.01a

%00.1001009000.4

09000.42a

%662.671000923.3

0000.10923.33a

The maximum absolute relative error after the first iteration is 100%


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Example

8118.3

7153.3

14679.0

3

2

1

x

x

x

After Iteration #1

14679.0

12

0923.359000.4311

x

7153.35

0923.3314679.0282 x

8118.3

13

900.4714679.03763

x

Substituting the x values into the equationsAfter Iteration #2

0923.3

9000.4

5000.0

3

2

1

x

x

x


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ExampleIteration #2 absolute relative approximate error%62.240100

14679.0

50000.014679.01

a

%887.311007153.3

9000.47153.32

a

%876.181008118.3

0923.38118.33

a

The maximum absolute relative error after the first iteration is 240.62%

This is much larger than the maximum absolute relative error obtained in

iteration #1. Is this a problem?


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ExampleRepeating more iterations, the following values are obtained1a

2a

3a

Iteration a

1 a

2 a

3

1

23

4

5

6

0.50000

0.146790.74275

0.94675

0.99177

0.99919

67.662

240.6280.23

21.547

4.5394

0.74260

4.900

3.71533.1644

3.0281

3.0034

3.0001

100.00

31.88717.409

4.5012

0.82240

0.11000

3.0923

3.81183.9708

3.9971

4.0001

4.0001

67.662

18.8764.0042

0.65798

0.07499

0.00000

4

3

1

3

2

1

x

x

x

0001.4

0001.3

99919.0

3

2

1

x

x

x

The solution obtained

is close to the exact solution of


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REFERENCES: http://numericalmethods.eng.usf.edu

http://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdf

Numerical Methods Lecture 2 Simultaneous Equations
http://numericalmethods.eng.usf.edu/http://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdfhttp://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdfhttp://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdfhttp://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdfhttp://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdfhttp://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdfhttp://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdfhttp://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdfhttp://www.essie.ufl.edu/~kgurl/Classes/Lect3421/NM2_matrix_s02.pdfhttp://numericalmethods.eng.usf.edu/

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